# Algebra: Solving for X (Data Sufficiency)

• Feb 12th 2011, 08:27 AM
dumluck
Algebra: Solving for X (Data Sufficiency)
Hi All,
Quick one on approach...

Question: What is a value of x?

(1) x^2 - 6 = -x
(2) x^2 = 4

This is how I tackled it...
(1) x^2 -x = 6
x = 6
(2) x^2 = 4
x = 6

So my answer would be that both (1) and (2) are sufficient to answer this question.

(1) x^2 - x - 6 = 0
x - 6 = 0
(x + 3)(x - 2) so x could be 3 or -2

(2) x^2 = 4
x^2 + 4 = 0
(x + 2)(x - 2)

Common across both is -2 so the answer would be a combination of 1 and 2.

My question is how does one know that they should set the equations equal to 0. Is it because the term 'solve' is used instead of.. 'What is x?'

Also I would assume that if either of the above solved to a perfect square that would be the answer?

• Feb 12th 2011, 08:34 AM
harish21
You are finding the unknown, which is x.

your approach to solving the problems is wrong.

\$\displaystyle x^2-6=-x\$

add 'x' on both sides of the equation, and you'll have:

\$\displaystyle x^2+x-6=0\$. Now solve for x using the appropriate approach.

and \$\displaystyle x^2=4\$ means \$\displaystyle x^2-4=0\$
• Feb 12th 2011, 08:46 AM
dumluck
Quote:

Originally Posted by harish21
You are finding the unknown, which is x.

your approach to solving the problems is wrong.

\$\displaystyle x^2-6=-x\$

add 'x' on both sides of the equation, and you'll have:

\$\displaystyle x^2+x-6=0\$. Now solve for x using the appropriate approach.

and \$\displaystyle x^2=4\$ means \$\displaystyle x^2-4=0\$

Thanks for the response. I know I'm wrong and I the answer makes sense once it is set to 0. My question was how I know to set the equation to 0 by the question; given that in instances where you are asked for x, when left with x^2 = 4. The natural tendency would be to say x = 2.
• Feb 12th 2011, 10:06 AM
Wilmer
Quote:

Originally Posted by dumluck
(1) x^2 - 6 = -x
(2) x^2 = 4

(1)x^2 + x - 6 = 0
(2)X^2 + 0 - 4 = 0
Subtract (2) from (1):
0 + x - 2 = 0
x = 2
Roger!
• Feb 13th 2011, 02:38 AM
HallsofIvy
Quote:

Originally Posted by dumluck
(1) x^2 - 6 = -x
(2) x^2 = 4

Add x to both sides of x^2- 6= -x: x^2- 6+ x= -x+ x so x^2+ x- 6= 0.

Subtract 4 from both side of x^2= 4: x^2- 4= 4- 4 so x^2- 4= 0.
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